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%I #12 Mar 31 2017 22:27:16
%S 1,2,8,14,27,94,137,284,947,1384,2847,9484,13847,28484,94847,138484,
%T 284847,948484,1384847,2848484,9484847,13848484,28484847,94848484,
%U 138484847,284848484,948484847,1384848484,2848484847,9484848484
%N a(n+1)=a(n)+a^(n), where the addition is in base 11 and where a^(n) is obtained from a(n) by replacing each digit with its multiplicative inverse modulo 11. Zero digits, if any, are deleted.
%C Conjecture. For any positive integer a(1), the sequence generated according to the above rule eventually cycles through the forms a(k)=[1][4^a][3][(84)^b],..., a(k+6)=[1][4^a][3][(84)^(b+1)], or through a(k)=[1][5^a][4][(84)^b],..., a(k+6)=[1][5^a][4][(84)^(b+1)], for nonnegative integers a and b. The sequence listed above, with a(1)=1, is an example of the first type.
%F Conjectures from _Colin Barker_, Feb 15 2016: (Start)
%F a(n) = a(n-2) + 10*a(n-3) - 10*a(n-5) for n>7.
%F G.f.: x*(1+2*x+7*x^2+2*x^3-x^4+10*x^5-10*x^6+10*x^8) / ((1-x)*(1+x)*(1-10*x^3)). (End)
%F From _Robert Israel_, Feb 15 2016: (Start)
%F For k >= 1:
%F a(6*k-4) = 2*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
%F a(6*k-3) = 94*10^(2*k-3) + 84*(10^(2*k-3)-10)/99 + 7.
%F a(6*k-2) = 13*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
%F a(6*k-1) = 2*10^(2*k-1) + 84*(10^(2*k-1)-10)/99 + 7.
%F a(6*k) = 94*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.
%F a(6*k+1) = 13*10^(2*k-1) + 84*(10^(2*k-1)-10)/99 + 7.
%F Colin Barker's conjectures follow from these. (End)
%K nonn,base,easy
%O 1,2
%A _John W. Layman_, Feb 14 2000
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